Question

Find the equation of the circle which passes through $(0,1),(1,0)$ and $(2,1)$.

Answer 1

Let us assume that the center of the circle is $C(h,k)$.

Now, $C$ is equidistant from $(0,1)$, $(1,0)$ and $(2,1)$.

$\therefore \sqrt{(h-0{)}^2+(k-1{)}^2}=\sqrt{(h-1{)}^2+(k-0{)}^2}$

$\therefore h^2+k^2-2k+1=h^2-2h+1+k^2$

$\therefore -2k=-2h$

$\therefore k=h\dots \left(1\right)$

Also, we have $\therefore \sqrt{(h-0{)}^2+(k-1{)}^2}=\sqrt{(h-2{)}^2+(k-1{)}^2}$

$\therefore (h-0{)}^2+(k-1{)}^2=(h-2{)}^2+(k-1{)}^2$

$\therefore h^2=(h-2{)}^2$

$\therefore h^2=h^2-4h+4$

$\therefore 4h=4$

$\therefore h=1\dots \left(2\right)$

From (1) and (2), $(h,k)=(1,1)$.

Hence, equation of circle with center $C$ and passing through $(0,1)$ is

$(x-h{)}^2+(y-k{)}^2=(0-h{)}^2+(1-k{)}^2$

$\therefore (x-1{)}^2+(y-1{)}^2=(0-1{)}^2+(1-1{)}^2$

$\therefore x^2-2x+1+y^2-2y+1=1$

$\therefore x^2+y^2-2x-2y+1=0$

This is the required equation of circle.